A diophantine equation of coprime integers

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I conjecture that for two coprime integers $a$ and $b$, for any integer $n$ coprime to $a$ and $b$ they exist integers $x$ and $y$ such that

  • $ax + by = n$
  • $a,b,x,y,n$ are pairwise coprime

I am having problems to prove it, specially with the second condition, although numerical experimentation seems to support it. Bezout's identity doesn't guarantee it, and the only cases I have been able to identify that for sure holds for any $a,b,x,y,n$ are $x=\pm 1$ and $y=\pm 1$ for $n=\pm (a\pm b)$. I believe it could be a good starting point for some kind of inductive argument, but I have not been able to make it work in a way that guarantees the second condition.

Any help or hint to prove this conjecture, or some counterexample to it, would be welcomed.

Thanks!

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We can assume without loss of generality that $\ n>0\ $, since if $\ a,b,x,y,n\ $ satisfy all the given conditions, then so do $\ a,$$\,b,$$\,{-}x,$$\,{-}y,$$\,{-}n\ $.

Using the extended Euclidean algorithm we can find integers $\ x_0,y_0\ $ such that $$ x_0a+ y_0b=1\ . $$ The integer $\ x_0\ $ must be relatively prime to $\ b\ $, and $\ y_0\ $ must be relatively prime to $\ a\ $. From the Chinese remainder theorem, it follows that there exists an integer $\ t\ $ such that \begin{align} t&\equiv -b^{-1}\pmod{n}\ \ , \ \ \ \ \text{ and}\\ t&\equiv b^{-1}(nx_0-1)\pmod{a}\ . \end{align} Let \begin{align} x=&nx_0-tb - kabn\ , \ \ \text{ and}\\ y=&ny_0+at+ka^2n\ \end{align} where $\ k $ is an integer which remains to be chosen. But whatever the value of $\ k\ $, $$ xa+yb=n\ , $$ $\ x\ $ is relatively prime to $\ b\ $, and $\ y\ $ is relatively prime to $\ a\ $, \begin{align} x&\equiv1\pmod{n}\ ,\text{ and}\\ x&\equiv1\pmod{a}\ ,\\ y&\equiv ny_0+at\equiv -b^{-1}a \pmod{n}\ . \end{align}and hence $\ x\ $ is relatively prime to $\ a\ $ and $\ n\ $ and $\ y\ $ is relatively prime to $\ n\ $. Since $\ a\ $ and $\ n\ $ are relatively prime to $\ b\ $ there is an integer $\ k\ $ such that $$ y=y_0+at+ka^2n\equiv1\pmod{b}\ . $$ Therefore, if we choose this $\ k\ $ in the equations for $\ x\ $ and $\ y\ $ above, then $\ y\ $ will be relatively prime to $\ b\ $ and the integers $\ a,b,x,y,n\ $ will satisfy all the given conditions.